The Texas Rangers have about a 1 in 4000 chance of making the playoffs

If you look at the playoff odds on FanGraphs.com right now, you’ll see the Texas Rangers listed as having a 0.0% chance of making the playoffs this year. But that doesn’t mean they have no chance. It just means their chance is so small that it doesn’t round up to 0.1%; instead it rounds down to 0.0%, as any chance less than 1 in 2000 will do. As it turns out, their chance of making the playoffs is about 1 in 4000 right now.

How we get to that number involves a lot of logical reasoning, complicated by the fact that the Rangers will play a series against one of the four teams they’re chasing, and there will be two series played this week between some of those same four teams.

Let’s set the stage properly. Here are the 8 remaining playoff contenders in the American League:

Only 6 teams in the American League may go to the playoffs. To be one of those 6, the Rangers must pass 2 of the 7 teams ahead of them in the standings (so long as one of them is not a division winner). Fortunately for the Rangers, there are 4 teams they still have a chance to reach. Unfortunately, they’ll be very difficult to reach.

Notice that if the Rangers win all 6 of their remaining games, and the Red Sox lose all 6 of theirs, that the Rangers would only manage to be tied with the Red Sox. But because they hold the tiebreaker over the Red Sox (having won 4 of the 7 games played between them this year), the Rangers would beat out the Red Sox in that case.

The same goes for Detroit. The Tigers must lose all 6 of theirs, and the Rangers must win all 6 of theirs, for the Rangers to tie; because they win the tiebreaker (having won 4 of 6 against the Tigers), the Rangers would beat out the Tigers.

The Rangers did not win their season series against the Astros, however, so must beat them by a game in the final standings, to pass them for a playoff spot. Because they are currently 5 games behind them, that could only happen if the Rangers win all 6 of their remaining games, and the Astros lose all 6 of theirs.

For the Rangers to catch the Guardians, they’ll have to win some of their remaining 3 games against them; those wins would give the tiebreaker to the Rangers. So the Rangers could stand to lose 1 game, or could stand the Guardians winning 1 game, and still beat them for a playoff spot.

Given that there’s only 1 team that isn’t forcing the Rangers to win all their remaining games, but that they need to beat at least 2 of these teams, the only option for the Rangers is to win all their remaining games.

Let’s start a list of requirements like this one:

We’re assuming here that every game a team plays the rest of the way has a 1/2 chance of being a win, and a 1/2 chance of being a loss. Because the Rangers have 6 games remaining, and there’s only 1 way to achieve the stated outcome (Rangers win all 6), that’s 1 outcome out of 26 possible outcomes, or a 1/64 chance of it happening.

What other outcomes must we consider?

Well if none of these teams were playing each other in these final 6 games, it would be a little less complicated. All the outcomes would be independent, so we could calculate the odds of each team’s win totals independently, as a starting point. Our list of requirements would look like this:

Because the Rangers would have to beat at least 2 of these teams, we’d take pairs of outcomes and calculate their odds:

[ (Red Sox lose all) AND (Tigers lose all) ]
OR [ (Red Sox lose all) AND (Astros lose all) ]
OR [ (Red Sox lose all) AND (Guardians lose 5 or 6) ]
OR [ (Tigers lose all) AND (Astros lose all) ]
OR [ (Astros lose all) AND (Guardians lose 5 or 6) ]

Notice that we didn’t include (Tigers lose all) AND (Guardians lose 5 or 6). That’s because one of those teams will win the central division; beating a division winner doesn’t help you win a wild card spot. They have to beat at least one of the Red Sox or Astros to get into the playoffs.

So we would multiply odds everywhere there’s an AND above, and then add them everywhere there is an OR above.

This would double-count or triple-count some cases though. For example, it would triple count the case where all three of these occur: (Red Sox lose all) AND (Tigers lose all) AND (Astros lose all). We’d have to subtract out double the odds of that happening.

After making a few more adjustments where 3 of those occur, we’d probably have one final adjustment to make for the case where all 4 occur:

(Red Sox lose all) AND (Tigers lose all) AND (Astros lose all) AND (Guardians lose 5 or 6).

Then we’d multiply our result by the odds of the Rangers winning all their games, because that has to happen in every case of the Rangers making the playoffs.

But we don’t live in that world. We live in a world where, in the final games of the season:

The Tigers play 3 games against the Red Sox
The Tigers play 3 games against the Guardians
The Rangers play 3 games against the Guardians

Oh my. This reduces the number of games that determine the Rangers’ fate from 30 down to 21. That’s good for the Rangers, because it means a lot fewer games would have to go a certain way for them to make the playoffs, and that gives them better odds.

It also changes how we do this. Now the outcomes we need to consider look like this:

I’ve used highlighting to show outcomes that are related to each other in that they cannot both happen. For example, looking at the two lines in gold, we see that the Red Sox cannot simultaneously lose all their remaining games while also winning all 3 against the Tigers.

Let’s consider those two middle lines right now. If the Tigers lose all their remaining games, that means both the Red Sox and Guardians win at least 3 games. And that means the Rangers can’t beat either of those teams. The only team left that they could beat is the Astros. So if the Rangers beat the Tigers, they must also beat the Astros (and only the Astros) to get into the playoffs. That gives us this:

(Tigers lose all) AND (Astros lose all)

Which is actually this:

(Red Sox win all 3 against the Tigers) AND (Guardians win all 3 against the Tigers) AND (Astros lose all)

And there is no chance of double-counting with other outcomes. This will very much simplify our work to compensate for double countings.

To this we add the following:

[ (Red Sox lose all) AND (Astros lose all) ]
OR [ (Red Sox lose all) AND (Guardians lose 5 or 6) ]
OR [ (Astros lose all) AND (Guardians lose 5 or 6) ]

But consider that in the end we’ll be multiplying everything by the odds of (Rangers win all), which must happen in every scenario. Because the Rangers play 3 of those games against the Guardians, that means three of the Guardian’s losses have already been accounted for by the (Rangers win all) outcome. So we only need to consider the additional chance that the Guardians lose 2 or 3 against the Tigers. So the above becomes:

[ (Red Sox lose all) AND (Astros lose all) ]
OR [ (Red Sox lose all) AND (Guardians lose 2 or 3 to Tigers) ]
OR [ (Astros lose all) AND (Guardians lose 2 or 3 to Tigers) ]

Notice that in all 3 of these scenarios, the Tigers become unreachable to the Rangers, because they win at least 2 games. The only double or triple counting in this trio of scenarios is where the Rangers beat everyone but the Tigers:

(Red Sox lose all) AND (Astros lose all) AND (Guardians lose 2 or 3 to Tigers)

That’s a triple-count, so we have to subtract double the odds of that happening.

We can put all this together, with odds, in a new chart:

We add the first four lines then subtract 2 times the last line to compensate for double counting:

2-12 + 2-12 + 2-7 + 2-7 – 2 x 2-13 = 2-6 + 2-12 = 65/4096.

Which we multiply by the odds of the Rangers winning all 6 of their remaining games, to give 65/262144. That’s about 1 in 4033, or 0.0248%.

Had it not been for teams playing each other, the odds would have been 1 in about 16,186. So the Ranger’s chances of making the playoffs are about 4 times better because of these teams playing against each other.

Nobody backed into this wildcard spot!

Wow, what a finish in the AL wildcard race.  The Indians winning their last 10 games, and needing every one of those wins to take the top wild card spot, as the Rays and Rangers both went 8-2 at the end (the Rangers with a 7-game winning streak of their own).  This is the way you want to see a playoff race finish … lots of winning!

Reminds me a bit of the Rockies’ mad dash to the playoffs at the end of the 2007 season.  They had to win, I believe, 13 of their last 14 to tie for the wildcard spot.  It was exciting to watch!

Predictions:  The Rays are a stong team, but weaker on the road.  The Rangers have been impressive how they’ve turned things around at the end of the season, turning a big slump into a big winning streak.  I’m inclined to give the Rangers the edge here given their home field advantage and momentum, but the fact that the Rays have David Price pitching gives me pause.  At times this season he’s been lights-out, and though lately he’s been just consistently good, I have to think for a big game like this you’ll see him pitching well.  Slight edge to the Rays because of Price.  Should Price get injured before he’s done, edge goes back to the Rangers.

As for the following matchup with the Indians, I give the edge to the Indians.  They’re a good team with a manager that knows about getting into and through the postseason, in Terry Francona.  They’re on a roll and they’ll have a very partisan crowd in their favor, as their Wednesday wildcard game is already sold out, and their opponent is yet to be determined.

But in the end, the ALCS will be between the Red Sox and the A’s.  Let’s just hope for a lot of fun baseball to watch on the way there!

Red Sox chances of having home-field advantage about 7 out of 11

The Red Sox will lose any tiebreaker against the Oakland A’s this season, should their records end up tied.  As each team has only 4 games remaining, and the Red Sox have just a 1 game lead over the A’s, to end up ahead of the A’s, the Red Sox must at least match the A’s win for win.

There are 256 possible outcomes of the Red Sox and A’s remaining 8 games (16 outcomes for the Red Sox’s 4 games times 16 outcomes for the A’s 4 games).  Outcomes here means, for example, win-loss-win-win, or win-win-win-loss (order matters).

When you examine all possible combinations of these outcomes, and even factor in their likelihoods of occurring, and keep in mind the statements made above, it turns out that the Red Sox have about a 7 out of 11 chance of securing home-field advantage over the A’s.  When you factor in the remote possibility of the Red Sox beating out Oakland for home-field advantage, but not beating out Detroit, the odds are slightly lower, but still about 7 out of 11.

Now let’s break down some of the above statements to see what’s behind them.

First, why is it that the Red Sox will lose any tiebreaker against the Oakland A’s?

The first tiebreaker is record in head-to-head games between the teams.  With each team winning 3 of the 6 games they’ve played against each other, that tiebreaker has no effect.

The next tiebreaker is intradivision record (record against the other teams within their own division).  Each team will finish with 76 intradivision games, and currently each team has 30 intradivision losses.  So this tiebreaker will go to the team that gets fewer intradivision losses the rest of the way.

This may seem hard to predict, but we can use two facts to our advantage here: 1) To end up tied, the Red Sox must lose exactly one more game the rest of the way than do the A’s. 2) There is only one non-intradivision game left for either team, and that is tonight’s Red Sox-Rockies game.

If the Red Sox win tonight’s game with the Rockies but end up tied with the A’s, it will be because they lost one more of their other games, all intradivision games, than did the A’s.  Since both teams currently have the same number of intradivision losses, that will give the Red Sox one more intradivision loss than the A’s, and the worse intradivision record, and so they lose this tiebreaker to the A’s.

If on the other hand the Red Sox lose tonight’s game with the Rockies but end up tied with the A’s, it will be because they lost the same number of their other games, all intradivision games, than did the A’s.  Both teams end up with the same intradivision record in this case, and so we move on to the next tiebreaker.

The next tiebreaker is higher winning percentage in the last half of intraleague games – games against other teams in the American League.  Currently, the Red Sox are 40-28 in these games, and the A’s are 40-27.  The Rockies game is again the only one of the remaining games that doesn’t contribute to this tiebreaker, which means, since we are considering the case in which the Red Sox and A’s lose the same number of the other games, all of which are intraleague games, the Red Sox would end up with one more loss in the same number of games for this tiebreaker, thus having the worse record, and losing the tiebreaker.

All of the above considered, the Red Sox will lose any tie with the A’s for best record, and therefore must maintain or grow their current one-game lead over the A’s to get home-field advantage.

Next we consider the question of why this means the Red Sox have a 7 out of 11 chance of getting home field advantage.

Of the 256 possible outcomes of the 8 remaining Red Sox and A’s games, in 163 of them the Red Sox at least match the A’s win for win, if not surpass them.  This represents 63.7% of the 256 possible outcomes.  If we assume all outcomes have an equal possibility of occurring, that means the Red Sox have a 63.7% chance of ending up with home field advantage over the A’s, or about 7 out of 11 (which is 63.64%).  But saying all outcomes have an equal possibility of occurring is to assume that both teams have a 50-50 chance of winning each of their remaining games, and that’s probably not the case.  If instead we assume a 60% chance for each team winning each remaining game (which essentially matches their winning percentages on the season), we can redo the calculation, weighting less likely outcomes (like loss-loss-loss-loss) lower than more likely outcomes (like win-win-win-win).  When you do this, the Red Sox’s odds turn out to be just a little bit better, 64.0%, to end up with home field advantage over the A’s.  Still pretty much 7 out of 11.

To get their chances of ending up with home field advantage, period, we have to subtract the likelihood of the Tigers tying or surpassing the Red Sox, while the A’s do not.  These odds, right now, are very small.  Assuming 50-50 chances in the games, the Red Sox have a 63.5% chance of ending up with home field advantage, and assuming 60-40 chances in the remaining games for each division leader, the Red Sox have a 63.9% chance.

No matter how you slice it, it’s pretty much 7 out of 11.

Of course you could argue that due to schedules, the odds are better now for one team or the other.  But I think the schedules are not too slanted for one team or the other right now, so I’m sticking with 7 out of 11.

I hope somebody out there enjoys reading this even half as much as I enjoyed producing it.

My sources for the data and tiebreaker information in this post were:

http://www.overthemonster.com/2013/9/23/4761626/red-sox-athletics-playoff-tiebreaker-home-field-advantage http://en.wikipedia.org/wiki/Major_League_Baseball_tie-breaking_procedures

Royals don’t have a shot

My last blog post is now no longer applicable.  With the Royals losing last night and everybody ahead of them winning, they’re out of it.  Don’t care that there’s still a mathematical chance for them; it’s too slim to heed.  We all can ignore Kansas City, Baltimore, and New York now, and just focus on the Rays, Indians, and Rangers.

I predict Rays and Indians.  (Yeah, a stretch, I know.)